Integrand size = 35, antiderivative size = 541 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {2 \left (7 a^2 A b^2-3 A b^4-4 a^3 b B+a^4 C+3 a^2 b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 a^2 (a-b) b^2 (a+b)^{3/2} d}+\frac {2 \left (a A b^2+3 A b^3+a^3 (3 B+C)-a^2 b (6 A+B+3 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 a^2 b \sqrt {a+b} \left (a^2-b^2\right ) d}-\frac {2 A \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a^3 d}+\frac {2 \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{3 a \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (3 A b^4+4 a^3 b B-a^4 C-a^2 b^2 (7 A+3 C)\right ) \tan (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \]
2/3*(7*A*a^2*b^2-3*A*b^4-4*B*a^3*b+C*a^4+3*C*a^2*b^2)*cot(d*x+c)*EllipticE ((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c)) /(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a^2/(a-b)/b^2/(a+b)^(3/2)/d+ 2/3*(a*A*b^2+3*A*b^3+a^3*(3*B+C)-a^2*b*(6*A+B+3*C))*cot(d*x+c)*EllipticF(( a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/( a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/a^2/b/(a^2-b^2)/d/(a+b)^(1/2)- 2*A*cot(d*x+c)*EllipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((a+b )/(a-b))^(1/2))*(a+b)^(1/2)*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+ c))/(a-b))^(1/2)/a^3/d+2/3*(A*b^2-a*(B*b-C*a))*tan(d*x+c)/a/(a^2-b^2)/d/(a +b*sec(d*x+c))^(3/2)-2/3*(3*A*b^4+4*B*a^3*b-a^4*C-a^2*b^2*(7*A+3*C))*tan(d *x+c)/a^2/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(11366\) vs. \(2(541)=1082\).
Time = 29.18 (sec) , antiderivative size = 11366, normalized size of antiderivative = 21.01 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Result too large to show} \]
Time = 2.17 (sec) , antiderivative size = 565, normalized size of antiderivative = 1.04, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4548, 27, 3042, 4548, 27, 3042, 4546, 3042, 4409, 3042, 4271, 4319, 4492}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4548 |
| \(\displaystyle \frac {2 \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}-\frac {2 \int -\frac {\left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x)-3 a (A b+C b-a B) \sec (c+d x)+3 A \left (a^2-b^2\right )}{2 (a+b \sec (c+d x))^{3/2}}dx}{3 a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x)-3 a (A b+C b-a B) \sec (c+d x)+3 A \left (a^2-b^2\right )}{(a+b \sec (c+d x))^{3/2}}dx}{3 a \left (a^2-b^2\right )}+\frac {2 \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (A b^2-a (b B-a C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-3 a (A b+C b-a B) \csc \left (c+d x+\frac {\pi }{2}\right )+3 A \left (a^2-b^2\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{3 a \left (a^2-b^2\right )}+\frac {2 \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4548 |
| \(\displaystyle \frac {-\frac {2 \int -\frac {3 A \left (a^2-b^2\right )^2+\left (-C a^4+4 b B a^3-b^2 (7 A+3 C) a^2+3 A b^4\right ) \sec ^2(c+d x)+a \left (3 B a^3-2 b (3 A+2 C) a^2+b^2 B a+2 A b^3\right ) \sec (c+d x)}{2 \sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \left (a^4 (-C)+4 a^3 b B-a^2 b^2 (7 A+3 C)+3 A b^4\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {3 A \left (a^2-b^2\right )^2+\left (-C a^4+4 b B a^3-b^2 (7 A+3 C) a^2+3 A b^4\right ) \sec ^2(c+d x)+a \left (3 B a^3-2 b (3 A+2 C) a^2+b^2 B a+2 A b^3\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \left (a^4 (-C)+4 a^3 b B-a^2 b^2 (7 A+3 C)+3 A b^4\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {3 A \left (a^2-b^2\right )^2+\left (-C a^4+4 b B a^3-b^2 (7 A+3 C) a^2+3 A b^4\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a \left (3 B a^3-2 b (3 A+2 C) a^2+b^2 B a+2 A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \left (a^4 (-C)+4 a^3 b B-a^2 b^2 (7 A+3 C)+3 A b^4\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4546 |
| \(\displaystyle \frac {\frac {\left (a^4 (-C)+4 a^3 b B-a^2 b^2 (7 A+3 C)+3 A b^4\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx+\int \frac {3 A \left (a^2-b^2\right )^2+\left (C a^4-4 b B a^3+b^2 (7 A+3 C) a^2+\left (3 B a^3-2 b (3 A+2 C) a^2+b^2 B a+2 A b^3\right ) a-3 A b^4\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{a \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \left (a^4 (-C)+4 a^3 b B-a^2 b^2 (7 A+3 C)+3 A b^4\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (a^4 (-C)+4 a^3 b B-a^2 b^2 (7 A+3 C)+3 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\int \frac {3 A \left (a^2-b^2\right )^2+\left (C a^4-4 b B a^3+b^2 (7 A+3 C) a^2+\left (3 B a^3-2 b (3 A+2 C) a^2+b^2 B a+2 A b^3\right ) a-3 A b^4\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \left (a^4 (-C)+4 a^3 b B-a^2 b^2 (7 A+3 C)+3 A b^4\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4409 |
| \(\displaystyle \frac {\frac {3 A \left (a^2-b^2\right )^2 \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx+(a-b) \left (a^3 (3 B+C)-a^2 b (6 A+B+3 C)+a A b^2+3 A b^3\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+\left (a^4 (-C)+4 a^3 b B-a^2 b^2 (7 A+3 C)+3 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \left (a^4 (-C)+4 a^3 b B-a^2 b^2 (7 A+3 C)+3 A b^4\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 A \left (a^2-b^2\right )^2 \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+(a-b) \left (a^3 (3 B+C)-a^2 b (6 A+B+3 C)+a A b^2+3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (a^4 (-C)+4 a^3 b B-a^2 b^2 (7 A+3 C)+3 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \left (a^4 (-C)+4 a^3 b B-a^2 b^2 (7 A+3 C)+3 A b^4\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4271 |
| \(\displaystyle \frac {\frac {(a-b) \left (a^3 (3 B+C)-a^2 b (6 A+B+3 C)+a A b^2+3 A b^3\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (a^4 (-C)+4 a^3 b B-a^2 b^2 (7 A+3 C)+3 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 A \sqrt {a+b} \left (a^2-b^2\right )^2 \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}}{a \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \left (a^4 (-C)+4 a^3 b B-a^2 b^2 (7 A+3 C)+3 A b^4\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4319 |
| \(\displaystyle \frac {\frac {\left (a^4 (-C)+4 a^3 b B-a^2 b^2 (7 A+3 C)+3 A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 A \sqrt {a+b} \left (a^2-b^2\right )^2 \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}+\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (a^3 (3 B+C)-a^2 b (6 A+B+3 C)+a A b^2+3 A b^3\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{a \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \left (a^4 (-C)+4 a^3 b B-a^2 b^2 (7 A+3 C)+3 A b^4\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}+\frac {2 \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}\) |
| \(\Big \downarrow \) 4492 |
| \(\displaystyle \frac {2 \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{3 a d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {\frac {-\frac {6 A \sqrt {a+b} \left (a^2-b^2\right )^2 \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{a d}+\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (a^3 (3 B+C)-a^2 b (6 A+B+3 C)+a A b^2+3 A b^3\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (a^4 (-C)+4 a^3 b B-a^2 b^2 (7 A+3 C)+3 A b^4\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}}{a \left (a^2-b^2\right )}-\frac {2 \tan (c+d x) \left (a^4 (-C)+4 a^3 b B-a^2 b^2 (7 A+3 C)+3 A b^4\right )}{a d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}}{3 a \left (a^2-b^2\right )}\) |
(2*(A*b^2 - a*(b*B - a*C))*Tan[c + d*x])/(3*a*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) + (((-2*(a - b)*Sqrt[a + b]*(3*A*b^4 + 4*a^3*b*B - a^4*C - a ^2*b^2*(7*A + 3*C))*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]] /Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[ -((b*(1 + Sec[c + d*x]))/(a - b))])/(b^2*d) + (2*(a - b)*Sqrt[a + b]*(a*A* b^2 + 3*A*b^3 + a^3*(3*B + C) - a^2*b*(6*A + B + 3*C))*Cot[c + d*x]*Ellipt icF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b *(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b* d) - (6*A*Sqrt[a + b]*(a^2 - b^2)^2*Cot[c + d*x]*EllipticPi[(a + b)/a, Arc Sin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - S ec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(a*d))/(a* (a^2 - b^2)) - (2*(3*A*b^4 + 4*a^3*b*B - a^4*C - a^2*b^2*(7*A + 3*C))*Tan[ c + d*x])/(a*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]))/(3*a*(a^2 - b^2))
3.10.73.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) *((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt [a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ .) + (a_)], x_Symbol] :> Simp[c Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[d Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Int[(A + (B - C )*Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]], x] + Simp[C Int[Csc[e + f*x]*(( 1 + Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A , B, C}, x] && NeQ[a^2 - b^2, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(a*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(a*(m + 1)*(a^2 - b^2)) Int[(a + b*Csc[e + f*x])^( m + 1)*Simp[A*(a^2 - b^2)*(m + 1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x ] + (A*b^2 - a*b*B + a^2*C)*(m + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(10171\) vs. \(2(502)=1004\).
Time = 17.13 (sec) , antiderivative size = 10172, normalized size of antiderivative = 18.80
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(10172\) |
| default | \(\text {Expression too large to display}\) | \(10322\) |
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integral((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c) + a)/ (b^3*sec(d*x + c)^3 + 3*a*b^2*sec(d*x + c)^2 + 3*a^2*b*sec(d*x + c) + a^3) , x)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]